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          Northern Eurasia Finals Online 2020
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        <p>2021第一场组队训练，手都有点生，Dirt比较高，慢慢回手感吧。</p>
<a id="more"></a>
<p><a href="https://codeforces.ml/gym/102896" target="_blank" rel="noopener" title=".com为原地址">比赛地址</a></p>
<h1 id="a-almost-balanced-tree">A Almost Balanced Tree</h1>
<h2 id="题意">题意</h2>
<p>你需要构造一棵 <span class="math inline">\(A+B\ (A+B\le10^5)\)</span> 个点的<strong>二叉树</strong>，其中 <span class="math inline">\(A\)</span> 个节点的点权值为 <span class="math inline">\(1\)</span>， <span class="math inline">\(B\)</span> 个节点的点权值为 <span class="math inline">\(2\)</span>。对于每个节点，它左子树的节点权值和与右子树的节点权值和相差不超过 <span class="math inline">\(1\)</span>。</p>
<p>如果可以构造，输出构造方案，否则输出 <span class="math inline">\(-1\)</span> 。</p>
<h2 id="题解">题解</h2>
<p>设 <span class="math inline">\(DP[i]\)</span> 表示：构造一颗合法的权值和为 <span class="math inline">\(i\)</span> 的树，最少需要多少权值为 <span class="math inline">\(1\)</span> 的点。一个权值为 <span class="math inline">\(2\)</span> 的点，可以拆分成两个权值为 <span class="math inline">\(1\)</span> 的点，一个留在原地，另一个分配到左右子树中任意一个来保持平衡。所以最少花费以上的都能构造出来。</p>
<p>那么转移是非常有限的，确定树的根是 <span class="math inline">\(1\)</span> 还是 <span class="math inline">\(2\)</span> 以后，由于平衡的限制，两个子树的权值和就也确定了。</p>
<p>比较麻烦的是，边转移要边构造，最后还要把多余的 <span class="math inline">\(2\)</span> 转为 <span class="math inline">\(1\)</span> ，胡神说每次转换都是log级别的，因为这种方法构造出来的树的深度就是log级别的，还有些细节不知道他是怎么弄好的。</p>
<h1 id="b-brain-teaser">B Brain-teaser</h1>
<h2 id="题意-1">题意</h2>
<table>
<thead>
<tr class="header">
<th></th>
<th style="text-align: center;"></th>
<th style="text-align: center;">S</th>
<th style="text-align: center;">E</th>
<th style="text-align: center;">N</th>
<th style="text-align: center;">D</th>
</tr>
</thead>
<tbody>
<tr class="odd">
<td>+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;">M</td>
<td style="text-align: center;">O</td>
<td style="text-align: center;">R</td>
<td style="text-align: center;">E</td>
</tr>
<tr class="even">
<td>=</td>
<td style="text-align: center;">M</td>
<td style="text-align: center;">O</td>
<td style="text-align: center;">N</td>
<td style="text-align: center;">E</td>
<td style="text-align: center;">Y</td>
</tr>
</tbody>
</table>
<p>一个形如上表格的竖式谜语，有一组解 <span class="math inline">\(O = 0\)</span> , <span class="math inline">\(M = 1\)</span> , <span class="math inline">\(Y = 2\)</span> , <span class="math inline">\(E = 5\)</span> , <span class="math inline">\(N = 6\)</span> , <span class="math inline">\(D = 7\)</span> , <span class="math inline">\(R = 8\)</span> , <span class="math inline">\(S = 9\)</span> 。</p>
<p>定义一个合法的谜语要求如下：</p>
<ul>
<li><p>每个字母代表一个确定的数字，且互不相同。</p></li>
<li><p>不存在任何前导 <span class="math inline">\(0\)</span> 。</p></li>
<li><p>谜语的解唯一。</p></li>
</ul>
<p>现在给出两个相加的字符串，再给出 <span class="math inline">\(n\)</span> 个字符串组成的字典，求字典中哪些字符串能作为第三个字符串，和给出的两个字符串组成合法的谜语。</p>
<p>特别地，首先给出的两个字符串以及字典中的字符串，长度均不少于 <span class="math inline">\(2\)</span> 且不超过 <span class="math inline">\(15\)</span> ，并且全部来自 <strong>Collins Dictionary Scrabble Words (2019)</strong>，共计279496个单词。</p>
<h2 id="题解-1">题解</h2>
<p>首先基本思路是暴力，先给已有的两个字符串枚举一个解，做出加法的结果，对结果中已经安排好的数字确定字母，这样就能得到一个模式串：其中一些字母是确定的，一些字母是待定的。</p>
<p>字典里的单词做出一颗字典树，再用这些模式串去匹配字典里的单词，进一步判断未安排的数字是否能够构成合法解。</p>
<p>但是这样复杂度似乎有些问题：一共有 <span class="math inline">\(10\)</span> 个数字,枚举解的数量是 <span class="math inline">\(10!\)</span> ，然后每个解要去匹配 <span class="math inline">\(n\)</span> 个单词，不考虑检测是否合法的情况下，总复杂度就已经达到 <span class="math inline">\(O(10!\times n)\)</span> ，会超时。</p>
<p>但是考虑到所有的单词都是人类使用的英语单词，并非为了卡到复杂度上限而特意构造的一些字符串，所以当我们枚举到一个解，并再字典树上匹配时，期望并不会匹配到很多合法的单词，所以复杂度期望是能够接受的。</p>
<p>然而写完后测试发现，所有模式串的总匹配量还是能达到百万级别，且匹配过程耗时就很大。</p>
<p>于是我将匹配后检测是否有合法解改为在字典树上边走边检测，这才从 <span class="math inline">\(TLE\)</span> 改成 <span class="math inline">\(AC\)</span> 。</p>
<h1 id="c-color-the-tree">C Color the Tree</h1>
<h2 id="题意-2">题意</h2>
<p>给出一颗 <span class="math inline">\(n\ (n\le20)\)</span> 个点的树，初始状态下，除去根节点 <span class="math inline">\(1\)</span> 是红色以外，其他所有的点都是白色的。</p>
<p>如果说树上每个红色的节点，从它到根节点路径上每个点都是红色的，那么这样的一个树的状态是合法的。</p>
<p>你可以选择树上除根节点外任意一个点，把它从红变白，或从白变红，要求操作前后树都是处于合法状态。</p>
<p>你需要求出一个最长的操作序列，使得整个操作过程都是合法的，且没有出现过重复的状态。</p>
<h2 id="题解-2">题解</h2>
<p>对于任意一个子树，我们都能做出把它从全白操作到已遍历所有合法状态的一个操作序列。构造方法如下：</p>
<ol type="1">
<li><p>先将子树的根变红，激活它所有儿子的子树。</p></li>
<li><p>先递归下去，把所有儿子的子树的操作序列构造出来。</p></li>
<li><p>依次将自己的儿子合并，合并两个儿子的操作如下：假设两个儿子的操作序列分别为 <span class="math inline">\(a_1,a_2,a_3,\dots a_n\)</span> 和 <span class="math inline">\(b_1,b_2,b_3,\dots b_m\)</span> ，那么合并后的序列为： <span class="math display">\[
a_1,a_2,a_3,\dots a_n,\ b_1,\ a_n,a_{n-1},\dots a_2,a_1,\ b_2,a_1,a_2,a_3,\dots a_n,\dots \dots,b_m,\ a_1,a_2,a_3,\dots a_n
\]</span></p>
<p>就是说，先正向遍历 <span class="math inline">\(a\)</span> 序列，再遍历 <span class="math inline">\(b\)</span> 序列中一个操作，再反向遍历 <span class="math inline">\(a\)</span> 序列，再遍历 <span class="math inline">\(b\)</span> 序列中一个操作，交替经行，直到 <span class="math inline">\(b\)</span> 序列被遍历完毕。</p></li>
</ol>
<p>这样就能把该子树所有的合法状态都遍历到。</p>
<p>总复杂度就是一颗树的合法状态数量，是 <span class="math inline">\(O(2^n)\)</span> 的。</p>
<h1 id="d-down-we-dig">D Down We Dig</h1>
<h2 id="题意-3">题意</h2>
<p><span class="math inline">\(Dina\)</span> 和 <span class="math inline">\(Dima\)</span> 在一个 <span class="math inline">\(n\ (n\le3\times10^5)\)</span> 行，<span class="math inline">\(8\)</span> 列的矩阵上博弈。</p>
<p>每一行都是一个长度为 <span class="math inline">\(8\)</span> 的，只包含 <span class="math inline">\(R\)</span> 和 <span class="math inline">\(W\)</span> 的字符串。</p>
<p>从第一行出发，双方轮流移动到更高行，能够移动的条件是：要从第 <span class="math inline">\(i\)</span> 行移动到第 <span class="math inline">\(j\)</span> 行，那么要满足这两行字符串上对应位置相等的字符数量 <span class="math inline">\(x\ge j-i\ge1\)</span>。</p>
<p>谁不能找到一个更高行移动，谁就失败。</p>
<p>输出 <span class="math inline">\(n\)</span> 个答案，第 <span class="math inline">\(i\)</span> 个答案表示当游戏只包含前 <span class="math inline">\(i\)</span> 行的时候，先手必胜还是后手必胜。</p>
<h2 id="题解-3">题解</h2>
<p>显然第，从一行最多跳 <span class="math inline">\(8\)</span> 行，因为相同的字符数量最多只有 <span class="math inline">\(8\)</span>。</p>
<p>这个游戏是单次无组合的，所以只要考虑必胜态和必败态。</p>
<p>一个 <span class="math inline">\(O(n^2)\)</span> 的思路就是每次询问从后往前算出每一行是必胜态还是必败态，但这么做肯定会超时。</p>
<p>不妨设置一个状压方程 <span class="math inline">\(DP[i][s]\)</span> 表示，当游戏进行到第 <span class="math inline">\(i\)</span> 行且之后 <span class="math inline">\(8\)</span> 行的胜败状态为 <span class="math inline">\(s\)</span> 的时候，先手是必胜还是必败，因为当我们拥有连续 <span class="math inline">\(8\)</span> 行的胜败状态后，我们就已经能够明确倒推出处于第一行时先手是否必胜了，这样的状态量最多是 <span class="math inline">\(O(2^8\times n)\)</span> 的，用记忆化就能做到这个复杂度。</p>
<p>那么我们每次询问就只要暴力算出最后 <span class="math inline">\(8\)</span> 行的胜败状态，就能开始用以上的状压DP的方法求解了。</p>
<h1 id="e-easy-measurements">E Easy Measurements</h1>
<h2 id="题意-4">题意</h2>
<p>两个水泵，第一个 <span class="math inline">\(b\)</span> 秒能泵 <span class="math inline">\(a\)</span> 升水，第二个 <span class="math inline">\(d\)</span> 秒能泵 <span class="math inline">\(c\)</span> 升水，这时发现两个一起上 <span class="math inline">\(d\)</span> 秒能泵 <span class="math inline">\(b\)</span> 升水。（ <span class="math inline">\(a,b,c,d\)</span> 均为正整数）</p>
<p>现在给出一组 <span class="math inline">\(b, d\ (1\le b,\ d\le10^9)\)</span> ，求有多少组选择 <span class="math inline">\(a,\ c\)</span> 的合法解。</p>
<h2 id="题解-4">题解</h2>
<p>转换一下：</p>
<ul>
<li><p>第一个 <span class="math inline">\(bd\)</span> 秒能泵 <span class="math inline">\(ad\)</span> 升水。</p></li>
<li><p>第二个 <span class="math inline">\(bd\)</span> 秒能泵 <span class="math inline">\(bc\)</span> 升水。</p></li>
<li><p>两个一起上 <span class="math inline">\(bd\)</span> 秒能泵 <span class="math inline">\(b^2\)</span> 升水。</p></li>
</ul>
<p>那么显然地：<span class="math inline">\(ad+bc=b^2\)</span>。转换一下可得： <span class="math display">\[
a = \frac{b(b-c)}{d}
\]</span> 此时约分一下 <span class="math inline">\(b\)</span> 和 <span class="math inline">\(d\)</span> 可以得到 <span class="math display">\[
a = \frac{b&#39;(b-c)}{d&#39;}
\]</span></p>
<p>要使得 <span class="math inline">\(a\)</span> 为正整数，那么 <span class="math display">\[
\frac{(b-c)}{d&#39;}
\]</span></p>
<p>就得是整数。</p>
<p>有多少 <span class="math inline">\(c\)</span> 满足条件呢，自然是： <span class="math display">\[
\lfloor\frac{b}{d&#39;}\rfloor
\]</span> 这就是答案，因为 <span class="math inline">\(a\)</span> 总是唯一确定且不同的，最后再根据所有数字都是正整数调整一下，正确答案为： <span class="math display">\[
\lfloor\frac{b-1}{d&#39;}\rfloor
\]</span></p>

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